Srib Interview Questions, Process, and Tips

cost price(c.p) of one fruit = 24/16
(c.p) = Rs 1.50
selling price(s.p) of one fruit = 18/8
(s.p) =Rs 2.25
profit for one fruit = 2.25 – 1.50 = 0.75
profit
profit % = ———– * 100
cost price
0.75
= ——- * 100
1.50
= 50%

160600

The ratio flat vs hill is unknown so you can’t calculate this normally. But since the average for uphill/downhill is also 4 kmph ((1.5h)/6km)), the calculation is: 6 hours * 4kmph = 24km.

CPNCBZ

Nice

Please explain to me why the answer is C. Thank you

Let water and milk be in x quantity
3x+2x=125
5x=125
x=25
Milk=3x=75litre
Water=2x=50litre
2x=50litre
4x=100litre

So 50 litre more should be added so that the ratio becomes 3:4

avg sal of 3 is $950
first person earns=$1150
second person earns=$650
let thirdperson =x
avg=(1150+650+x)/3
950*3=1800+x
x=2850-1800
x=$1050

1200 meter

Traffic Jams
Public Speaking
What to Wear to a Party
Noises on an Airplane
Bees

q

In the given series 5,6,7,8,10,11,14…
There are two series
First is 5,7,10,14…
Second is 6,8,11….
First series
5+2=7
7+3=10
10+4=14
Second series
6+2=8
8+3=11
11+4=15
Hence complete series is 5,6,7,8,10,11,14,15…

To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.

The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.

There are 24 primes in S, so the product of S is:

2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97

We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.

5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.

25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.

Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros